Question

Use the t-distribution to find a confidence interval for a mean
μ given the relevant sample results. Give the best point estimate
for μ, the margin of error, and the confidence interval. Assume the
results come from a random sample from a population that is
approximately normally distributed.

A 90% confidence interval for μ using the sample results x¯=142.0,
s=52.8, and n=50

Round your answer for the point estimate to one decimal place, and
your answers for the margin of error and the confidence interval to
two decimal places.

point estimate =

margin of error =

The 90% confidence interval is to .

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 142.0

sample standard deviation = s = 52.8

sample size = n = 50

Degrees of freedom = df = n - 1 = 49

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t_{
/2,df} = t_{0.05,49} = 1.677

Margin of error = E = t_{/2,df}
* (s /n)

= 1.677 * (52.8 / 50)

= 12.52

The 90% confidence interval estimate of the population mean is,

- E < < + E

142.0 - 12.52 < < 142.0 + 12.52

**129.48 <
< 154.52**

**point estimate = 142.0
margin of error = 12.52
The 90% confidence interval is 129.48 to 154.52**

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